In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?
We use complementary counting. There are a total of $3^5=243$ ways to color the five squares without any restrictions. Now we consider the case where there are fewer than three yellow squares.
If there are no yellow squares, then each of the five squares must be either red or blue. It is easy to see that there are $2^5=32$ ways to color the squares in this case.
If there is exactly one yellow square, then each of the other four squares must be either red or blue. This gives $2^4=16$ ways to color the squares.
Finally, if there are exactly two yellow squares, then the squares must alternate in color, either in the pattern YRYRY or YBRYB. Each case gives $2$ colorings. (Note that any other patterns of two yellow squares would result in at least two consecutive squares of the same color.) So there are a total of $2+2=4$ ways to color the squares in this case.
Therefore, there are $32+16+4= \boxed{52}$ colorings in which there are fewer than three yellow squares. The answer is $243-52=\boxed{191}$.